Let $M_1, M_2, ... M_k$ be smooth manifolds with or without boundary such that at most one of these has nonempty boundary, say $M_a$ has nonempty boundary where $a$ is in $\{1,2, ..., k\}$.

Then it is said that $M_1 \times M_2 \times ... \times M_k$ is a smooth manifold with boundary. I am curious what its boundary is.

Also, does the same contents hold for just topological manifolds? I am somewhat confused...

  • $\begingroup$ See also here. $\endgroup$ – Jan Bohr Mar 4 '18 at 18:34

In both cases the boundary is $X= M_1\times\dots\times\partial M_a \times\dots\times M_k$.

My post was downvoted, so perhaps I missed the intention of the question? The reason the boundary is as above is because any point not in $X$ is contained in a coordinate neighborhood given by $(U,\phi)=(U_1\times\dots\times U_k,\phi_1\times\dots\times\phi_k)$, where $(U_i,\phi_i)$ is an interior coordinate chart for each $i$. This neighborhood is thus an interior coordinate chart for $M$, which follows because the product of open subsets $\mathbb{R}^{n_i}$ is an open subset of $\mathbb{R}^{n_1+\dots +n_k}$.

For a point $p=(p_1,\dots,p_k)$ in $X$, the corresponding $(U,\phi)$ is a product of an open subset of $\mathbb{H}^{n_a}$ and open subsets of $\mathbb{R}^{n_i}$, $i\ne a$. This is an open subset of $\mathbb{H}^{n_1+\dots+n_k}$, with $p$ in the boundary.

So we have precisely $\partial M=X$. A similar argument works for topological manifolds.


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